with 1.0 mL of .50 m KI and collect and dry the CuI precipitate

you mix 6.0 mL of .50 M Cu(NO3)2 with 1.0 mL of .50 m KI and collect and dry the CuI precipitate. Calculate the following:moles of Cu(NO3)2 used:moles of KI used:moles of CuI precipitate expected from KI:What mass of CuI do you expect to obtain from this reaction?If the mass of the precipitate you recover is .038g, what is the percent recovery of the precipitate?

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